Finding sum of a geometric series

Question

I am asked to find the summation of $1/3^n$ from $n=5$ to infinity.

I have done the calculation: $1/(1-r)$, for $r=1/3$, and received $1.5$. As this summation starts from $5$, I subtracted $3^0, 3^-1, 3^-2, 3^-3$ and $3^-4$ from $1.5$ and got $6.17e-3$. However, apparently this answer is wrong, and so is the answer 0. I appreciate any help, thank you!

Answer 1

$$1,\frac{1}{3},\frac{1}{3^2},\frac{1}{3^3},\frac{1}{3^4} \ldots$$

You need to calculate $$S_{\infty}-S_5$$

$$=\frac{1}{1-\frac{1}{3}}-\frac{1(1-\frac{1}{3^5})}{1-\frac{1}{3}}$$

$$=\frac{\frac{1}{3^5}}{\frac{2}{3}}$$

$$=\frac{1}{2\cdot{3^4}}$$

$$=\frac{1}{162}$$

Answer 2

Your method is fine indeed

$$\sum_{k=5}^\infty \frac1{3^k}=\sum_{k=0}^\infty \frac1{3^k}-\sum_{k=0}^4 \frac1{3^k}=\frac32 - \sum_{k=0}^4 \frac1{3^k}=\frac32-1-\frac13-\frac19-\frac1{27}-\frac1{81}=\frac1{162}$$

As an alternative, following the clever method suggested by lulu in the comment, we have

$$\sum_{k=5}^\infty \frac1{3^k}=\sum_{j=0}^\infty \frac1{3^{j+5}}=\frac1{3^5}\sum_{j=0}^\infty \frac1{3^{j}}=\frac1{3^5}\cdot\frac32=\frac1{2\cdot 3^4}=\frac1{162}$$

Answer 3

You also can use this formula, valide if $\;|x|<1$: $$\sum_{n=N}^\infty x^n=\frac{x^N}{1-x},$$ which is a straightforward corollary of the formula for the sum of the geometric series: $$\sum_{n=N}^\infty x^n=x^N\sum_{n=0}^\infty x^n=x^N\frac1{1-x}.$$